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我有如下矩阵；从计算中会有一些值在图中绘制时彼此具有相同的值（我使用散点图），绘制时的冗余值 这里的问题是如何知道/识别哪个计算给出了匹配数据并且可以将它存储在桌子。我曾尝试在 matlab 中使用 find 函数，但出现错误。

A = [2/3 -1/3 -1/3;
    -1/3 2/3 -1/3;
    -1/3 -1/3 2/3];
B=[0 0 0;
    0 0 1;
    0 1 0;
    0 1 1;
    1 0 0;
    1 0 1;
    1 1 0;
    1 1 1];
d= [2/3 -1/3 -1/3];
q = [0 0.5773 -0.5773];
for i = 1:size(B,1)
    p1 =  A * B(1,:)' -A * B(i,:)';
    dtransformation0a(i) = d*p1;
    qtransformation0a(i) = q*p1;
    p2 =  A * B(2,:)' -A * B(i,:)'
    dtransformation0b(i) = d*p2;
    qtransformation0b(i) = q*p2;
     p3 =  A * B(3,:)' -A * B(i,:)';
    dtransformation0c(i) = d*p3;
    qtransformation0c(i) = q*p3;
    p4 =  A * B(4,:)' -A * B(i,:)';
    dtransformation0d(i) = d*p4;
    qtransformation0d(i) = q*p4;
    p5 =  A * B(5,:)' -A * B(i,:)';
    dtransformation0e(i) = d*p5;
    qtransformation0e(i) = q*p5;
    p6 =  A * B(6,:)' -A * B(i,:)';
    dtransformation0f(i) = d*p6;
    qtransformation0f(i) = q*p6;
end
figure
 scatter(dtransformation0a,qtransformation0a,100,'b.')
hold on
scatter(dtransformation0b,qtransformation0b,100,'b.')
hold on
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1 回答 1

0

我现在无法访问 MATLAB 解释器，所以我无法自己测试它，但这应该可以：

A = [2/3 -1/3 -1/3;
    -1/3 2/3 -1/3;
    -1/3 -1/3 2/3];
B = [0 0 0;
    0 0 1;
    0 1 0;
    0 1 1;
    1 0 0;
    1 0 1;
    1 1 0;
    1 1 1];
d = [2/3 -1/3 -1/3];
q = [0 0.5773 -0.5773];
for i = 1:size(B,1)
    p1 =  A * B(1,:)' -A * B(i,:)';
    dtransformation0a(i) = d*p1;
    qtransformation0a(i) = q*p1;
    p2 =  A * B(2,:)' -A * B(i,:)'
    dtransformation0b(i) = d*p2;
    qtransformation0b(i) = q*p2;
     p3 =  A * B(3,:)' -A * B(i,:)';
    dtransformation0c(i) = d*p3;
    qtransformation0c(i) = q*p3;
    p4 =  A * B(4,:)' -A * B(i,:)';
    dtransformation0d(i) = d*p4;
    qtransformation0d(i) = q*p4;
    p5 =  A * B(5,:)' -A * B(i,:)';
    dtransformation0e(i) = d*p5;
    qtransformation0e(i) = q*p5;
    p6 =  A * B(6,:)' -A * B(i,:)';
    dtransformation0f(i) = d*p6;
    qtransformation0f(i) = q*p6;
end

X = [dtransformation0a, dtransformation0b];    %Assuming dtransformation0a and dtransformation0b are row martices.
Y = [qtransformation0a, qtransformation0b];    %Assuming qtransformation0a and qtransformation0b are row martices.
my_points = [X';Y'];    %Each row of my_points will represent a particular point.
my_points = unique(my_points, 'rows');    %Get unique rows, hence points.

figure
scatter(my_points(:,1),my_points(:,2),100,'b.')

如果您遇到任何问题，请告诉我。

**

编辑！！

**

根据您的要求，我的理解是您想要显示那些产生值的计算，例如(0,0)，如果这确实是您想要的，您应该能够通过将循环部分替换为：

my_point = [0, 0];
for i = 1:size(B,1)
    p1 =  A * B(1,:)' -A * B(i,:)';
    dtransformation0a(i) = d*p1;
    qtransformation0a(i) = q*p1;
    p2 =  A * B(2,:)' -A * B(i,:)'
    dtransformation0b(i) = d*p2;
    qtransformation0b(i) = q*p2;
     p3 =  A * B(3,:)' -A * B(i,:)';
    dtransformation0c(i) = d*p3;
    qtransformation0c(i) = q*p3;
    p4 =  A * B(4,:)' -A * B(i,:)';
    dtransformation0d(i) = d*p4;
    qtransformation0d(i) = q*p4;
    p5 =  A * B(5,:)' -A * B(i,:)';
    dtransformation0e(i) = d*p5;
    qtransformation0e(i) = q*p5;
    p6 =  A * B(6,:)' -A * B(i,:)';
    dtransformation0f(i) = d*p6;
    qtransformation0f(i) = q*p6;

    if(dtransformation0a(i)==my_point(1))
        fprintf("dtransformation0a => (%f = %f * %f;)\n",dtransformation0a(i),d,p1);
    end
    if(dtransformation0b(i)==my_point(1))
        fprintf("dtransformation0b => (%f = %f * %f;)\n",dtransformation0b(i),d,p2);
    end
    if(qtransformation0a(i)==my_point(2))
        fprintf("qtransformation0a => (%f = %f * %f;)\n",qtransformation0a(i),q,p1);
    end
    if(qtransformation0b(i)==my_point(2))
        fprintf("qtransformation0b => (%f = %f * %f;)\n",qtransformation0b(i),q,p2);
    end
end

我希望这就是你要找的。

于 2013-03-12T04:13:25.030 回答